Question

The area of the circle $x^2+y^2=16$ exterior to the parabola $y^2=6x$ is

• A. $\frac{4}{3}(4\pi -\sqrt{3})$
• B. $\frac{4}{3}(4\pi +\sqrt{3})$
• C. $\frac{4}{3}(8\pi -\sqrt{3})$
• D. $\frac{4}{3}(8\pi +\sqrt{3})$

Answer 1

The correct option is C $\frac{4}{3}(8\pi -\sqrt{3})$

The given equations are
$x^2+y^2=\mathrm{16........}\left(1\right)$
$y^2=6x.......\left(2\right)$

Area bounded by the circle and parabola
$=2\left[Area\right(OADO)+Area(ADBA\left)\right]$

$=2[{\int }_0^2\sqrt{16x}dx+{\int }_2^4\sqrt{16-x^2}dx]$

$=2\left[\sqrt{6}{\left\{\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right\}}_0^2\right]+2{[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}]}_2^4$

$=2\sqrt{6}\times \frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_0^2+2[8\cdot \frac{\pi }{2}-\sqrt{16-4}-8{\sin }^{-1}\left(\frac{1}{2}\right)]$

$=\frac{4\sqrt{6}}{3}\left(2\sqrt{2}\right)+2[4\pi -\sqrt{12}-8\frac{\pi }{6}]$

$=\frac{16\sqrt{3}}{3}+8\pi -4\sqrt{3}-\frac{8}{3}\pi$

$=\frac{4}{3}[4\sqrt{3}+6\pi -2\sqrt{3}-2\pi ]$$=\frac{4}{3}[\sqrt{3}+4\pi ]$

$=\frac{4}{3}[4\pi +\sqrt{3}]$ sq. units

Area of circle $=\pi (r{)}^2$
$=\pi (4{)}^2$
$=16\pi$ sq. units

$\therefore$ Required area $=16\pi -\frac{4}{3}[4\pi +\sqrt{3}]$
$=\frac{4}{3}[4\times 3\pi -4\pi -\sqrt{3}]$

$=\frac{4}{3}(8\pi -\sqrt{3})$ sq. units

Thus, the correct answer is C.